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(A) Given,$f(	heta) = \sin \left(	an^{-1} \left(\frac{\sin 	heta}{\sqrt{\cos 2	heta}} ight) ight)$.We know that $\cos 2	heta = \cos^2 	heta

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(A) Given,$f(	heta) = \sin \left(	an^{-1} \left(\frac{\sin 	heta}{\sqrt{\cos 2	heta}} ight) ight)$.We know that $\cos 2	heta = \cos^2 	heta - \sin^2 	heta = 1 - 2\sin^2 	heta$.Let $	an^{-1} \left(\frac{\sin 	heta}{\sqrt{\cos 2	heta}} ight) = \alpha$. Then $	an \alpha = \frac{\sin 	heta}{\sqrt{\cos 2	heta}}$.Using the identity $1 + 	an^2 \alpha = \sec^2 \alpha$,we have $\sec^2 \alpha = 1 + \frac{\sin^2 	heta}{\cos 2	heta} = \frac{\cos 2	heta + \sin^2 	heta}{\cos 2	heta} = \frac{\cos^2 	heta - \sin^2 	heta + \sin^2 	heta}{\cos 2	heta} = \frac{\cos^2 	heta}{\cos 2	heta}$.Thus,$\cos^2 \alpha = \frac{\cos 2	heta}{\cos^2 	heta}$,which implies $\sin^2 \alpha = 1 - \frac{\cos 2	heta}{\cos^2 	heta} = \frac{\cos^2 	heta - \cos 2	heta}{\cos^2 	heta} = \frac{\cos^2 	heta - (2\cos^2 	heta - 1)}{\cos^2 	heta} = \frac{1 - \cos^2 	heta}{\cos^2 	heta} = \frac{\sin^2 	heta}{\cos^2 	heta} = 	an^2 	heta$.Since $-\frac{\pi}{4} Therefore,$f(	heta) = \sin \alpha = 	an 	heta$.Now,$\frac{d}{d(	an 	heta)}(	an 	heta) = 1$.

Let $f(\theta) = \sin \left(\tan^{-1} \left(\frac{\sin \theta}{\sqrt{\cos 2\theta}} \right) \right)$,where $-\frac{\pi}{4} < \theta < \frac{\pi}{4}$. Then the value of $\frac{d}{d(\tan \theta)}(f(\theta))$ is

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